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Chapter 14: Problem 53

Consider the following equilibrium system involving \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (sulfuryl dichloride): $$\mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)$$ Predict how the equilibrium position would change if (a) \(\mathrm{Cl}_{2}\) gas were added to the system; (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) were removed from the system; (c) \(\mathrm{SO}_{2}\) were removed from the system. The temperature remains constant.

Short Answer

Expert verified
(a) The equilibrium would shift to the right when Cl₂ gas is added to the system. (b) The equilibrium also shifts to the right when SO₂Cl₂ is removed. (c) The equilibrium would shift to the left when SO₂ is removed from the system.

Step by step solution

01

Analyze addition of Cl₂

According to Le Chatelier's principle, when a reactant is added to the system, the equilibrium will shift to the right to re-establish equilibrium. Therefore, when Cl₂ gas is added to the system, the reaction will proceed in the direction of forming more SO₂Cl₂. This means that the position of the equilibrium would move to the right.
02

Analyze removal of SO₂Cl₂

Similarly, when a product is removed from the system, the reaction will shift towards the products side to make up for the lost quantity. In this case, when SO₂Cl₂ is removed, the reaction will move in the direction of forming more SO₂Cl₂ to restore balance. Thus, the equilibrium will shift towards the right.
03

Analyze removal of SO₂

However, when a reactant is removed, in this case SO₂, the reaction will shift towards the reactants side to compensate the lost quantity. Therefore, the system would react by reducing the amount of SO₂Cl₂ being produced and increase the production of SO₂ through backward reaction. Hence, the equilibrium would shift towards the left.

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Most popular questions from this chapter

Define equilibrium. Give two examples of a dynamic equilibrium.

Consider the reaction $$\begin{array}{r}2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) \\\\\Delta H^{\circ}=-198.2 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Comment on the changes in the concentrations of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at equilibrium if we were to (a) increase the temperature; (b) increase the pressure; (c) increase \(\mathrm{SO}_{2}\); (d) add a catalyst; (e) add helium at constant volume.

In the uncatalyzed reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ the pressure of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=\) 0.377 and \(P_{\mathrm{NO}_{2}}=1.56 \mathrm{~atm}\) at \(100^{\circ} \mathrm{C} .\) What would happen to these pressures if a catalyst were added to the mixture?

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.487 \mathrm{M} .\) The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) is \(4.63 \times 10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color. \(]\)

What is the law of mass action?
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